Problem: Simplify and expand the following expression: $ \dfrac{t - 4}{4t - 8}-\dfrac{t + 2}{t - 6} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(4t - 8)(t - 6)$ Multiply the first term by $\dfrac{t - 6}{t - 6}$ $ \begin{align*} \dfrac{t - 4}{4t - 8} \times \dfrac{t - 6}{t - 6} & = \dfrac{(t - 4)(t - 6)}{(4t - 8)(t - 6)} \\ & = \dfrac{t^2 - 10t + 24}{(4t - 8)(t - 6)}\end{align*} $ Multiply the second term by $\dfrac{4t - 8}{4t - 8}$ $ \begin{align*} \dfrac{t + 2}{t - 6} \times \dfrac{4t - 8}{4t - 8} & = \dfrac{(t + 2)(4t - 8)}{(t - 6)(4t - 8)} \\ & = \dfrac{4t^2 - 16}{(t - 6)(4t - 8)}\end{align*} $ Now we have: $ = \dfrac{t^2 - 10t + 24}{(4t - 8)(t - 6)} - \dfrac{4t^2 - 16}{(t - 6)(4t - 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{t^2 - 10t + 24 - (4t^2 - 16)}{(4t - 8)(t - 6)} $ $ = \dfrac{t^2 - 10t + 24 - 4t^2 + 16}{(4t - 8)(t - 6)} $ $ = \dfrac{-3t^2 - 10t + 40}{(4t - 8)(t - 6)}$ Expand the denominator: $ = \dfrac{-3t^2 - 10t + 40}{4t^2 - 32t + 48}$